Practice questions · Topic 4.5

Unit 4 · Practice for the Topic 4.5 end-of-topic test

These are practice questions in the shape of the topic test. Work through them before you take the test; every question tells you what it wanted.

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback tells you what a wrong choice assumed. For the free-response questions, write your answer in full sentences and show any calculation. The first free-response question walks you through the reasoning one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it. The percent of cells in a stage is the count in that stage divided by the total counted, times 100; the time a stage takes is that percent, as a fraction, times the length of the whole cycle, which the question gives. Where a graph's caption says its error bars represent ±2SE, the overlap rule applies.

Question 1

A strawberry plant sends out a long runner along the ground. At the runner's tip a small new plant forms, with its own roots and leaves, and it carries the same DNA as the parent plant.

How does the new plant come to exist?

A. Cells at the runner's tip swell until they are large enough to form roots and leaves B. Cells at the runner's tip divide again and again, each into two daughter cells C. Cells from the soil enter the runner's tip and grow into roots and leaves D. A seed inside the runner's tip sprouts into the new plant

Question 2

Between divisions, a stained cell from a rice root shows an evenly grainy nucleus and no separate rods.

Why is the cell's DNA in this form between divisions?

A. Grainy DNA is DNA that has been broken into short pieces ready for copying B. The rods have already been handed on to the daughter cells C. The chromosomes have coiled up tightly so they can be moved D. Each chromosome is spread out as chromatin, open to be read and copied

Question 3

A rabbit body cell has 44 chromosomes. In S phase it copies its DNA.

How many chromatids does the cell hold at the end of S phase?

A. 22 B. 44 C. 88 D. 176

Question 4

A cell in a growing fern root has just finished cytokinesis and become two daughter cells.

Which three stages does each daughter cell pass through next, in order?

A. G1, S phase, G2 B. S phase, G1, G2 C. G2, S phase, G1 D. Mitosis, G1, S phase

Question 5

Cells of a newt hold 18 pg of DNA in G1.

Which DNA reading shows a newt cell part-way through S phase?

A. 9 pg B. 18 pg C. 27 pg D. 36 pg

Question 6

When a salamander loses a leg, cells near the stump that last divided years earlier begin to copy their DNA within a few days, and soon they divide.

Which statement describes these cells before the injury and after it?

A. Before: held in G2 with copied DNA; after: straight into mitosis B. Before: in S phase, copying slowly; after: copying quickly C. Before: held in mitosis for years; after: finishing the division D. Before: in G0, working; after: back into G1 and on into S phase

Question 7

A corn cell with 20 chromosomes has lined them up at the equator, each still two joined sister chromatids, and now enters anaphase.

How many chromatids arrive at each pole?

A. 10 B. 20 C. 40 D. 80

Question 8

Two nuclei have formed in a dividing cell from a moss leaf, and the cytoplasm is now being divided.

Which structure divides the cytoplasm of this cell, and how is it made?

A. A cell plate, made by vesicles fusing at the equator into a new wall B. A cleavage furrow, made by a ring of protein under the membrane tightening C. A nuclear envelope, made by the spindle fibers joining up across the middle D. A centromere, made by the sister chromatids joining at the middle of the cell

Question 9

One cell from a frog embryo, fixed and stained; each body cell of this frog holds four chromosomes.

The drawing below shows one cell from a frog embryo, fixed and stained as it was dividing.

Which stage of mitosis does the drawing show?

A. Prophase B. Metaphase C. Anaphase D. Telophase

Question 10

Mean percent of cells in mitosis in the root tip and in a young leaf of the same bean seedling; five students each counted 200 cells of each tissue. Error bars represent ±2SE. Gridlines every 2%.

Five students each counted 200 cells from the root tip and 200 cells from a young leaf of the same bean seedling, recording how many were in mitosis. The graph below shows the mean percent of cells in mitosis for each tissue.

What do these data support?

A. Every root-tip cell divides faster than every leaf cell B. The leaf holds fewer cells in total than the root tip C. The two tissues have exactly the same share of dividing cells D. More of the root-tip cells are dividing than of the leaf cells

How to tackle the free-response questions. Read the verb first: describe asks what you see or know; explain asks why or how, so name the mechanism; predict asks what will happen and why; justify asks for the evidence that supports a claim. Each point is earned by one idea, stated in a sentence that names the thing and the mechanism. Extra words earn nothing; a wrong extra can lose the point. If there is a figure or table, use what it shows. When you finish, check the box for each point your answer earns and compare your sentences with the full-credit answer.

Free response 1 · Scientific Investigation · 5 points

A Hydra is a small freshwater animal that reproduces by budding: a new animal grows out from the side of its body and later drops off. A student asks whether more of the cells in a bud are dividing than in the rest of the body. Five students each fix and stain one budding Hydra, cut a thin slice through the bud and a second slice through the body column halfway between the mouth and the foot, and record the stage of 200 cells in each slice. The table gives one student's counts for her bud slice. The graph gives the mean percent of cells in mitosis for the five bud slices and the five body-column slices, with error bars that represent ±2SE. Take the whole cell cycle in a Hydra as 24 hours.

Top: the stage of each of the 200 cells the student counted on her slide of the bud. Bottom: the mean percent of cells in mitosis for the five students' bud slides and five body-column slides; error bars represent ±2SE. Gridlines every 5%.

(a) Identify the dependent variable in this investigation. (1 point)

Hint: Which quantity did each student read off her slides and record, as against the thing the students chose to compare?

A full-credit answer: The dependent variable is the percent of the 200 cells in each slice that were in mitosis, the quantity the students measured.

Check the box for each point your answer earns

the dependent variable is the percent of cells in mitosis (or the stage each of the 200 cells was recorded in).

Accept "the number of cells in mitosis out of 200". Do not award the point for "the bud and the body column" (that is the independent variable) or for "200 cells" (a condition kept the same).

Common slip: Naming the part of the Hydra (bud or body column). That is what the students chose to compare, the independent variable; the dependent variable is what they counted.

(b) Calculate the percent of the cells on the bud slide that were in mitosis. (1 point)

Hint: Which four stages in the table together make up mitosis, and what total does the percent equation divide their sum by?

Write down the values in the question:

cells in mitosis = 20 + 9 + 9 + 6 = 44

total cells counted = 200

Write down the equation:

percent in mitosis = \frac{cells in mitosis}{total cells counted} \times 100

Substitute the values into the equation:

percent in mitosis = \frac{cells in mitosis}{total cells counted} \times 100

percent in mitosis = \frac{44}{200} \times 100

percent in mitosis = 0.22 \times 100 = 22 %

A full-credit answer: On the bud slide 22% of the cells were in mitosis: 44 of the 200 cells were in prophase, metaphase, anaphase or telophase.

Do not award the point for 10% (prophase alone), 78% (the interphase share) or 28% (44 divided by 156, the interphase count).

(c) Calculate the time, in hours, that a bud cell spends in metaphase, taking the whole cycle as 24 hours. (1 point)

Hint: Start from the share of the 200 cells that were in metaphase. What does the question give as the length of the whole cycle, and how does the time-in-stage equation use that length?

Write down the values in the question:

cells in metaphase = 9

total cells counted = 200

cycle length = 24 h

Write down the equations:

percent in metaphase = \frac{cells in metaphase}{total cells counted} \times 100

time in metaphase = \frac{percent in metaphase}{100} \times cycle length

Substitute the values into the equations:

percent in metaphase = \frac{9}{200} \times 100 = 4.5 %

time in metaphase = \frac{4.5}{100} \times 24 h

time in metaphase = 0.045 \times 24 h = 1.08 h

1.08 h \times 60 min/h = 64.8 min \approx 65 min

A full-credit answer: A bud cell spends 1.08 hours in metaphase, about 65 minutes.

Accept 1.1 h. Do not award the point for 4.5 h (the percent read as hours), 0.045 h (the fraction with no cycle length) or 2.16 h (9 divided by 100 instead of 200).

(d) Justify the claim that more of the bud's cells are dividing than of the body column's cells, using the error bars. (1 point)

Hint: What does the caption say each bar represents, and what does the overlap rule say about two bars of that kind?

A full-credit answer: The caption says the bars represent ±2SE, so the overlap rule applies. The bud's bar runs from 19.0% to 25.0% of cells in mitosis and the body column's from 2.4% to 5.6%. The bars do not overlap, so the difference between the means is very unlikely to be chance: more of the bud's cells were caught in mitosis, so more of them are dividing.

Check the box for each point your answer earns

the ±2SE bar for the bud runs from 19.0% to 25.0% and the bar for the body column from 2.4% to 5.6%; the bars do not overlap, so the higher percent in mitosis in the bud is very unlikely to be chance, and more of the bud's cells were caught dividing.

Accept a comparison that quotes both ranges and says they are apart. Do not award the point for comparing the two means alone (22.0% against 4.0%).

Common slip: Comparing 22.0% with 4.0% and stopping. The claim rests on the ±2SE bars, which do not overlap; two means alone could differ by chance.

(e) Describe what a greater percent of cells in mitosis shows about the bud's cells, and what it leaves open. (1 point)

Hint: One slide freezes every cell at one instant. Think about what a count of frozen cells can tell you directly, and what about a single cell's cycle it cannot.

A full-credit answer: A greater percent of cells in mitosis shows that more of the bud's cells are in the act of dividing (or that mitosis takes longer in the bud). On its own it leaves open whether any one bud cell cycles faster than a body-column cell: a count of frozen cells shows how many were caught dividing, and the speed of a single cell would need the cycle length.

Check the box for each point your answer earns

a greater percent in mitosis shows that more of the bud's cells are dividing (or that mitosis takes longer there); it leaves open whether any single cell cycles faster, because the count cannot tell the two apart without the cycle length.

Accept any correct statement of what one frozen slide cannot show: whether any single cell cycles faster, the length of the cycle, or whether the daughter cells keep dividing. Do not award the point for "every bud cell divides faster" stated as what the data show.

Common slip: Turning a bigger share into "every bud cell divides faster". The count says how many of the sampled cells were dividing; it says nothing about the speed of any one cell.

Free-response score: 0 of 5

Free response 2 · Conceptual Analysis · 4 points

The drawing shows two cells from the root of Haplopappus, a small desert daisy whose body cells hold four chromosomes: two long and two short. The cells were fixed and stained at one instant and are labeled cell A and cell B.

Two cells from the root of Haplopappus, a small desert daisy whose body cells hold four chromosomes (two long and two short), fixed and stained at one instant.

(a) Identify the stage of mitosis shown by cell A, and describe the feature of the drawing that shows it. (1 point)

A full-credit answer: Cell A is in anaphase. The sister chromatids have separated, and two groups of four single chromatids are moving apart toward opposite poles with their centromeres leading, pulled by fibers from the centrosomes; no nuclear envelope surrounds them.

Check the box for each point your answer earns

anaphase, shown by two groups of single chromatids (V shapes with the centromere leading) moving apart toward opposite poles, pulled by the spindle fibers, with no envelope around them.

Accept "the sisters have separated and are moving to the poles". Do not award the point for metaphase (there is no row at the equator) or for telophase (the fibers are still attached and no envelope is forming).

Common slip: Naming telophase because the chromosomes are in two groups. In telophase the fibers are gone and an envelope forms around each group; here the fibers are still pulling and the groups are still on the move.

(b) Calculate the number of chromatids cell A held at metaphase, before its sister chromatids separated. (1 point)

Write down the values in the question:

chromosomes = 4

chromatids in each copied chromosome = 2

Write down the equation:

chromatids at metaphase = 2 \times chromosomes

Substitute the values into the equation:

chromatids at metaphase = 2 \times chromosomes

chromatids at metaphase = 2 \times 4 = 8

A full-credit answer: At metaphase cell A held 8 chromatids: each of its four chromosomes was two sister chromatids joined at a centromere, the eight single chromatids now moving to the poles.

Do not award the point for 4 (the chromosome count) or 16 (doubling twice).

(c) Explain why cell B is dividing its cytoplasm with a plate across its middle rather than with a groove pinching in. (1 point)

A full-credit answer: Cell B is a plant cell with a rigid cell wall, and a wall cannot be pinched in the way an animal cell's membrane can. So vesicles gather at the equator and fuse into a cell plate, the structure across the middle of the drawing, which grows outward until it meets the old wall and becomes the new wall between the two daughter cells.

Check the box for each point your answer earns

a plant cell's rigid cell wall cannot be pinched in, so vesicles gather at the equator and fuse into a cell plate that grows outward into a new wall between the two daughter cells.

Accept "the wall cannot be pinched, so a new wall is built between the nuclei". Do not award the point for "plant cells have no ring of protein" with no reference to the wall, or for describing the plate as the spindle.

Common slip: Saying the plate is the spindle left behind. The spindle broke down in telophase; the plate is new wall, built from vesicles fusing at the equator.

(d) Justify the claim that the two cells formed from cell B carry the same genome as each other. (1 point)

A full-credit answer: Before cell B divided, every one of its four chromosomes was copied once in S phase into two identical sister chromatids. In anaphase the sisters separated and the spindle sent one chromatid of every pair to each pole, so each of the two nuclei in cell B holds one copy of every chromosome. The two daughter cells therefore carry the same complete genome, four chromosomes each, identical to each other and to the parent.

Check the box for each point your answer earns

every chromosome was copied once in S phase into two identical sister chromatids, and anaphase sent one chromatid of every pair to each pole, so each daughter cell receives one complete copy of the genome: one of each of the four chromosomes, identical to the other daughter's.

Accept "each nucleus holds one copy of every chromosome because each was copied and the copies were separated". Do not award the point for "the parent shared its four chromosomes out, two to each side".

Common slip: Saying the four chromosomes were shared out two and two. The chromosomes were doubled first, in S phase, so each daughter cell receives all four.

Free-response score: 0 of 4

Multiple choice checked: 0 of 10 correct.

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